Solve for x and y:\(\frac{3}{x+y}+\frac{2}{x-y}=2\),\(\frac{9}{x+y}+\f

1.	Solve for x and y:\(\frac{3}{x+y}+\frac{2}{x-y}=2\),\(\frac{9}{x+y}+\frac{4}{x-y}=1\)
Answer» The given equations are \(\frac{3}{x+y}+\frac{2}{x-y}=2\)....(i) \(\frac{9}{x+y}+\frac{4}{x-y}=1\)......(ii) Substituting 1/x+y = u and 1/x−y = v, we get: 3u + 2v = 2 ……..(iii) 9u - 4v = 1 …….(iv) On multiplying (iii) by 2, we get: 6u + 4v = 4 …..(v) On adding (iv) and (v), we get: 15u = 5 ⇒u = 5/15 = 1/3 ⇒ 1/x+y = 1/3 ⇒ x + y = 3 …….(vi) On substituting u = 1/3 in (iii), we get 1 + 2v = 2 ⇒2v = 1 ⇒v = 1/2 ⇒ 1/x−y = 1/2 ⇒ x – y = 2 …….(vii) On adding (vi) and (vii), we get 2x = 5 ⇒ x = 5/2 On substituting x = 5/2 in (vi), we have 5/2 + y = 3 ⇒ y = \((3-\frac{5}2)\)= 1/2 Hence, the required solution is x = 5/2 and y = 1/2 .

Solve for x and y:\(\frac{3}{x+y}+\frac{2}{x-y}=2\),\(\frac{9}{x+y}+\frac{4}{x-y}=1\)

Answer»

The given equations are

\(\frac{3}{x+y}+\frac{2}{x-y}=2\)....(i)

\(\frac{9}{x+y}+\frac{4}{x-y}=1\)......(ii)

Substituting 1/x+y = u and 1/x−y = v, we get:

3u + 2v = 2 ……..(iii)

9u - 4v = 1 …….(iv)

On multiplying (iii) by 2, we get:

6u + 4v = 4 …..(v)

On adding (iv) and (v), we get:

15u = 5

⇒u = 5/15 = 1/3

⇒ 1/x+y = 1/3

⇒ x + y = 3 …….(vi)

On substituting u = 1/3 in (iii), we get

1 + 2v = 2

⇒2v = 1

⇒v = 1/2

⇒ 1/x−y = 1/2 ⇒ x – y = 2 …….(vii)

On adding (vi) and (vii), we get

2x = 5

⇒ x = 5/2

On substituting x = 5/2 in (vi), we have

5/2 + y = 3

⇒ y = \((3-\frac{5}2)\)= 1/2

Hence, the required solution is x = 5/2 and y = 1/2 .